BKL: suspension, uniform coefficients and KLS - The KLS theorem and its methods

Part of the Bizeul–Klartag–Lehec proof, Chapter Bizeul–Klartag–Lehec: cumulants and suspension; the reading order is on the full proofs page.

Author. bkl_suspension_author, gpt-6-astra, 2026-10-06 (researcher).

Overview. We reconstruct Bizeul et al., 2026, Section 7. A log-concave suspension places an arbitrary smooth affine-orthogonal test in one additional coordinate. Replicating the original coordinates makes the curvature cost arbitrarily small. A cumulant contraction then bounds the Taylor tensor. We supply the density argument, treat the affine component, and extend the resulting coefficient bound to every covariance contraction, including measures supported on proper subspaces. The spectral criterion is used only after the coefficient estimate has been established.

Dependencies. The conventions are Definition 8.1. The suspension implication itself uses no spectral-gap theorem. Its application uses Theorem 8.2. Approximation with convergence of all polynomial moments and scalar Poincaré stability use Lemma 8.1. The final KLS conclusion uses Theorem 8.1. None of the Song–Zhang KLS theorem, the exponential-coefficient equivalence, CMH, or an occupation hypothesis is used. These are exact interfaces; their proof dossiers require independent certification along with this dossier.

Theorem 135.1 (Suspension implication)

Fix an integer $d\ge2$ and $b>0$ . Suppose that every isotropic log-concave probability $\nu$ in every finite dimension satisfies $R_{d+1}^{\nu}\preceq bI$ . Then, for every regular isotropic log-concave probability $\mu$ and every real $f\in L^2(\mu)$ ,

|\mathcal T_d^\mu f|^2\le \frac{2b}{(d!)^2}\|f\|_2^2.

(135.1)

Here $F_f(z)=\int fe^{\langle z,x\rangle}\,d\mu(x)/ \int e^{\langle z,x\rangle}\,d\mu(x)$ , $\mathcal T_df=\nabla^dF_f(0)/d!$ , and $\langle R_k^\nu v,v\rangle=|\kappa_k^\nu(v,\cdot,\ldots,\cdot)|^2$ . All tensor norms sum over all ordered indices. This proves the implication Proposition 8.2, without asserting its premise on its own.

Proof 135.1

Write $d\mu=e^{-V}dx$ and $D^2V\succeq aI$ with $a>0$ . Initially suppose $f$ is smooth, has bounded Hessian, and satisfies

\int f\,d\mu=0,\qquad \int f^2\,d\mu=1,\qquad \int xf(x)\,d\mu(x)=0.

(135.2)

For $\beta>0$ , take independent $X_1,\ldots,X_N$ of law $\mu$ and an independent real variable $\eta_\beta$ of density $(\beta/2)e^{-\beta|t|}$ . Put

F_N=N^{-1/2}\sum_i f(X_i),\qquad \sigma_\beta^2=1+2\beta^{-2},\qquad S=(F_N+\eta_\beta)/\sigma_\beta.

(135.3)

The law $\nu$ of $(S,X_1,\ldots,X_N)$ has potential, up to a constant,

W(s,x_1,\ldots,x_N)=\sum_iV(x_i)+ \beta\left|\sigma_\beta s-N^{-1/2}\sum_i f(x_i)\right|.

(135.4)

Choose an integer $N$ so large that $\beta N^{-1/2}\sup_x\|D^2f(x)\|_{\rm op}\le a$ . The displayed potential is the maximum of the two convex functions

\sum_i\bigl(V(x_i)\pm\beta N^{-1/2}f(x_i)\bigr) \mp\beta\sigma_\beta s.

(135.5)

Thus $\nu$ is log-concave, although regularity of $\nu$ is not claimed or needed. Independence and the three normalizations give $\mathbb ES=0$ , $\mathbb ES^2=1$ , $\mathbb ESX_i=0$ , and $\operatorname{Cov}(X_i,X_j)=\delta_{ij}I$ . Consequently $\nu$ is isotropic.

Bounded Hessian gives $|f(x)|\le C(1+|x|^2)$ ; strong convexity gives $V(x)\ge a|x|^2/2-C(1+|x|)$ . Hence the joint Laplace transform is finite in a neighborhood of the origin: the coefficient of each quadratic term from $w f/(\sigma_\beta\sqrt N)$ is absorbed by $a|x|^2/2$ , and the Laplace variable of $\eta_\beta$ has modulus less than $\beta$ there. These bounds, with slightly enlarged parameters, dominate every finite number of derivatives, so differentiation under the integrals is valid. For the logarithmic Laplace transform $\Lambda$ of $\nu$ , independence and $\mathbb E\eta_\beta=0$ give

\partial_w\Lambda(0,z_1,\ldots,z_N) =\frac1{\sigma_\beta\sqrt N}\sum_{i=1}^N F_f(z_i).

(135.6)

Let index 0 denote $S$ , and $(i,j)$ denote coordinate $j$ of $X_i$ . Differentiating $d$ more times yields

\kappa_{d+1}^\nu{}_{0,(i_1,j_1),\ldots,(i_d,j_d)} =\begin{cases} \dfrac{d!}{\sigma_\beta\sqrt N}(\mathcal T_df)_{j_1,\ldots,j_d}, &i_1=\cdots=i_d,\\ 0,&\text{otherwise}. \end{cases}

(135.7)

The $N$ indicated tensor blocks have disjoint ordered-index supports. Entries involving another index 0 form an orthogonal residual tensor $U$ , so

|\kappa_{d+1}^\nu(e_0)|^2 =\frac{(d!)^2}{\sigma_\beta^2N}\sum_{i=1}^N|\mathcal T_df|^2+|U|^2 \ge\frac{(d!)^2}{\sigma_\beta^2}|\mathcal T_df|^2.

(135.8)

The assumed bound in dimension $Nn+1$ makes the left side at most $b$ . This holds for each $\beta$ , with a possibly different $N$ , and the quantity $\mathcal T_df$ does not depend on either parameter. Taking the infimum over $\beta>0$ proves $|\mathcal T_df|\le\sqrt b/d!$ . No limit of the suspension measures is taken.

We justify passage to all affine-orthogonal tests. For each fixed ordered index $J$ of length $d$ , differentiation of the quotient at zero writes $(\mathcal T_df)_J=\int f p_J\,d\mu$ , where $p_J$ is a polynomial of degree at most $d$ whose coefficients are finite combinations of moments of $\mu$ . Gaussian tails imply $p_J\in L^2(\mu)$ . Thus $\mathcal T_d:L^2(\mu)\to(\mathbb R^n)^{\otimes d}$ is continuous at this fixed measure and degree, with no uniform bound assumed.

Let $h$ be a unit vector in the orthogonal complement of the affine functions. Choose $g_j\in C_c^\infty$ with $g_j\to h$ in $L^2(\mu)$ . This density follows by first truncating the function and its support, then approximating on compact sets, where the smooth positive density is bounded above and below. The orthogonal projection onto the affine functions is explicitly

Pg=\int g\,d\mu+\left\langle\int xg(x)\,d\mu(x),x\right\rangle.

(135.9)

It is bounded since $1,x_1,\ldots,x_n$ are orthonormal. Hence $g_j-Pg_j\to h$ , its norm tends to one, and its Hessian is the Hessian of $g_j$ , therefore bounded. Normalize these functions for all sufficiently large $j$ and pass through the continuous map $\mathcal T_d$ . We obtain $|\mathcal T_dh|\le\sqrt b/d!$ , and by homogeneity the same bound multiplied by $\|h\|_2$ for every affine-orthogonal $h$ .

For a linear function $\ell(x)=\langle v,x\rangle$ , isotropy gives $\|\ell\|_2=|v|$ , and

F_\ell(z)=\langle v,\nabla\Lambda_\mu(z)\rangle, \qquad \mathcal T_d\ell=\kappa_{d+1}^\mu(v)/d!.

(135.10)

The same cumulant premise gives $|\mathcal T_d\ell|\le\sqrt b\|\ell\|_2/d!$ . For $f=c+\ell+h$ its orthogonal affine decomposition, constants have $\mathcal T_dc=0$ and consequently

|\mathcal T_df|\le\frac{\sqrt b}{d!}(\|\ell\|_2+\|h\|_2) \le\frac{\sqrt{2b}}{d!}\sqrt{\|\ell\|_2^2+\|h\|_2^2} \le\frac{\sqrt{2b}}{d!}\|f\|_2.

(135.11)

This completes the suspension implication.

Theorem 135.2 (Uniform coefficients, including singular covariance contractions)

Let $K\ge1$ be the universal constant in Theorem 8.2, so $R_{d+1}^\nu\preceq K^d(d!)^2I$ for every isotropic log-concave $\nu$ and every integer $d\ge2$ . Set $A=\sqrt{2K}$ . For every centered log-concave probability $\mu$ in every dimension with $\operatorname{Cov}(\mu)\preceq I$ , every integer $d\ge1$ , and every $f\in L^2(\mu)$ ,

|\mathcal T_d^\mu f|\le A^d\|f\|_2.

(135.12)

Equivalently the symmetric Appell coefficient obeys $c_d(\mu)=\sqrt{K_d(\mu)}/d!\le A^d$ , with $K_d(\mu)=\sup_{T=T^{\rm sym},\,|T|=1} \operatorname{Var}_\mu P_d^\mu[T]$ . This is Theorem 8.3.

Proof 135.2

|\mathcal T_1f|=\sup_{|v|=1}\left|\int\langle v,x\rangle f\,d\mu\right| \le\|f\|_2

(135.13)

by isotropy and Cauchy–Schwarz.

We give the precise duality needed to pass through approximation. In a neighborhood of zero the Appell generating identity is

e^{\langle z,x\rangle-\Lambda_\mu(z)} =\sum_{d\ge0}\frac1{d!}\langle P_d^\mu(x),z^{\otimes d}\rangle.

(135.14)

Differentiating at zero and integrating shows $\mathbb EP_d^\mu=0$ for $d\ge1$ , and

\langle\mathcal T_df,T\rangle =\frac1{d!}\int fP_d^\mu[T]\,d\mu.

(135.15)

All entries are polynomial moment identities; hence they also define the same tensors without any complex-analytic convention. Taking first the supremum over unit $f\in L^2(\mu)$ and then over unit symmetric $T$ shows

\|\mathcal T_d\|_{L^2\to\mathrm{HS}}^2 =\frac1{(d!)^2}\sup_{|T|=1,\,T=T^{\rm sym}} \int(P_d^\mu[T])^2\,d\mu=c_d(\mu)^2.

(135.16)

Restricting to symmetric tensors loses nothing because $\mathcal T_df$ is symmetric and orthogonal projection onto that subspace decreases norm.

Now take an arbitrary isotropic log-concave $\mu$ and regular isotropic $\mu_j$ converging weakly and in every polynomial moment, furnished by Lemma 8.1. For fixed $d$ and $T$ , $P_d^{\mu_j}[T]$ is a polynomial in $x$ of degree at most $d$ whose coefficients are polynomial expressions in moments of order at most $d$ . This follows directly by formal inversion of the Laplace series, whose constant term is one. Its coefficients therefore converge to those of $P_d^\mu[T]$ . The square has degree at most $2d$ ; moment convergence then gives

\int(P_d^\mu[T])^2\,d\mu =\lim_j\int(P_d^{\mu_j}[T])^2\,d\mu_j \le(d!)^2A^{2d}|T|^2.

(135.17)

This is for every $T$ , with the same constant. Taking its supremum gives $c_d(\mu)\le A^d$ . No convergence of suprema or of varying test functions is asserted. Duality returns the bound for every $f\in L^2(\mu)$ . For such $f$ its tilted integral is indeed defined near zero: a log-concave law on its linear support has an exponential moment in some neighborhood of zero, and Cauchy–Schwarz controls $\int |f|e^{\langle z,x\rangle}d\mu$ by $\|f\|_2(\int e^{2\langle z,x\rangle}d\mu)^{1/2}$ . The same estimate with polynomial factors justifies its finite derivatives.

Finally let $C=\operatorname{Cov}(\mu)\preceq I$ be possibly singular and put $E=\operatorname{ran}C$ . Centering implies $X\in E$ almost surely: for each $v\in\ker C$ , $\mathbb E\langle v,X\rangle^2=0$ . If $E=\{0\}$ , the law is a point mass and all positive-degree Taylor coefficients vanish. Otherwise $C_E$ is positive definite on $E$ , and $Y=C_E^{-1/2}X$ is isotropic log-concave on $E$ . Write $B:E\to\mathbb R^n$ for the inclusion composed with $C_E^{1/2}$ . Then $X=BY$ and $\|B\|_{\rm op}\le1$ . For $g(y)=f(By)$ , $\|g\|_{L^2(Y)}=\|f\|_{L^2(X)}$ and

F_f^\mu(z)=F_g^{\mathcal L(Y)}(B^*z),\qquad \mathcal T_d^\mu f=B^{\otimes d}\mathcal T_d^{\mathcal L(Y)}g.

(135.18)

The ordered-index Hilbert–Schmidt tensor norm satisfies $\|B^{\otimes d}\|_{\rm op}=\|B\|_{\rm op}^d\le1$ (for example, diagonalize $B^*B$ on $E$ ). This proves the claimed estimate and then its equivalent coefficient bound for every covariance contraction. All arguments up to this point are independent of KLS.

Proof 135.3

For a regular isotropic $\mu$ , the preceding theorem supplies simultaneously in every degree the hypothesis of Theorem 8.1, with $R=A=\sqrt{2K}$ . If $C$ is the universal constant of that criterion, it gives $C_P(\mu)\le CA^2=2CK$ .

For arbitrary isotropic log-concave $\mu$ , take regular isotropic $\mu_j\Rightarrow\mu$ as above. For fixed $q\in C_c^\infty$ , the functions $q$ , $q^2$ and $|\nabla q|^2$ are bounded continuous, so their integrals converge and yield $\operatorname{Var}_\mu q\le2CK\int|\nabla q|^2d\mu$ . The scalar stability clause of Lemma 8.1 extends this inequality to every locally Lipschitz finite-energy $f$ , including its $L^2$ integrability. The infinite-energy case has no finite upper-bound assertion. Thus $C_*=2CK$ is universal.

Fences respected. These nodes have no additional bounded_by edges. The circularity fence Remark 32.6 is respected because no moving-competitor profile is used. The ceiling distinction Remark 33.6 is respected: no covariance-time converse is asserted. The projection barrier Remark 25.4 is respected because the suspension controls the full ordered-index tensor norm, not only diagonal directional evaluations. Remark 33.5 is not used or upgraded. No CMH, sharp gate-zero, occupation, or Song–Zhang bounded-loss antecedent is discharged by this proof.

References¶

Bizeul, P., Klartag, B., & Lehec, J. (2026). Presenting a Proof of the Kannan–Lovasz–Simonovits Conjecture. https://arxiv.org/abs/2610.05474v1